Proportional representation

A simple method for proportional allocation of seats

I’m not sure which of the established methods (Jefferson, D’Hondt, Sainte-LaguĂ«) this corresponds to, but it’s a pretty easy-to-understand system.

First there must be an initial standing of seats. Start by allocating every party which passed the hurdle one seat, or else giving one seat to each party that got more than (100//n/)% of the vote, where n is the number of seats to be allocated (this is the Hare quota, but you could also use Droop). There are other ways to get an initial standing, such as in mixed-member proportional representation, where the initial standing is the number of local seats won by each party.

Now we must calculate each party’s relative under- or over-representation. This is simply the percentage of seats held by each party according to the current standing, minus their percentage of the vote. (This is an overly simplistic method which allocates a lot of seats to the larger parties to start with, then moves on to the smaller parties. A better formula would ensure the allocation is balanced at each individual stage of the process.)

Now, until all seats are filled, repetitively give a seat to the party with the worst under-representation (that is, the lowest value of the misrepresentation function) then recalculate the relative under- and over-representations. If two or more parties are equally badly represented, give one seat to each of them and then recalculate.

A worked example

In this hypothetical election, there are 13 seats to give out (I picked this number because it’s small but not too small, and prime but not too close to the number 10 to make the result obvious) and the parties got the following results:

  • Red Party: 32.6% (907 votes)
  • Orange Party: 29.8% (830 votes)
  • Blue Party: 22.5% (626 votes)
  • Green Party: 10.5% (292 votes)
  • Yellow Party: 4.6% (127 votes)

The Yellow Party didn’t reach the 5% threshold and is eliminated from the race. All the other parties got more than 1/13th of the vote (obviously), so to start our algorithm off, all the other parties get one seat each. This means each eligible party has 7.69% of the available seats. Their misrepresentations are: Red -24.91%; Orange -22.14%; Blue -14.81%; Green -2.80%.

Since Red is the most under-represented, we give them the first extra seat. This changes their misrepresentation to -17.22%, so now the Orange party is the most unfairly under-represented. They get one seat bringing them to -14.45%, making Red once again the most under-represented by the current standing.

Giving Red their third seat makes Blue the most under-represented, and giving Blue their second seat makes Orange the worst under-represented. This continues until all 13 seats have been given out: Red; Blue; Orange; Green.

The final standing of the parties is as following:

  • Red: 4 seats, misrepresentation -1.83%
  • Orange: 4 seats, misrepresentation +0.93%
  • Blue: 3 seats, misrepresentation +0.58%
  • Green: 2 seats, misrepresentation +4.89%
  • (Yellow: 0 seats, misrepresentation -4.57%)

In a real election with many seats (national legislatures typically have hundreds), the result would give Red a few more seats than Orange, establishing their plurality. As it is, Orange could lead government with Blue even though they didn’t get the largest number of votes (though still a majority together). There’d also be a smaller over-representation of Green at the end. For instance, in a house with 131 seats, the result would be: Red 44 (+0.99%); Orange 41 (+1.46%); Blue 31 (+1.16%); Green 15 (+0.95%); Yellow 0 (-4.57%).